Ghasem D. Najafpour. Biochemical Engineering and Biotechnology
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234 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
Solution
Basis: 1 day operation
Cellulose Balance
Cellulose in the wastewater stream is not generated but it is consumed and biomass
produced by the microbial population as a substrate for energy.
[Cellulose in] – [cellulose out] [cellulose generation] [cellulose consumption]
[cellulose accumulation]
Biomass Balance
Applying the same concept for mass balance, the mass of sludge accumulated is:
Cell in – cell out growth rate death rate cell accumulation
9.5 CONSERVATION OF MASS PRINCIPLE
Based on the law of conservation of mass, mass is neither created nor destroyed, except in
nuclear reactions. Therefore, in a definite system, mass in is equal to mass out.
Example 1
Humid air with O
2
is prepared for a gluconic acid fermentation. The humid air has been pre-
pared by a special humidification chamber where 1.5 l/h of liquid water enters; at the same
10 1.5 10 2 10 5 10 5,500 kg bacteria
In Out Generation
3442
Consumption Accumulated cells
10 10 0 7 10 20,000 kg cellulose
In Out Generation Cons
54 4
uumption Accumulated cellulose
TABLE 9.3. Density and mass fraction of inlet and outlet of the fermenter
Substance Density, kg/m
3
Inlet (kg) Outlet (kg) Mass fraction
Inlet Outlet
Glucose 1544 1670.40 0.00 0.80 —
H
2
SO
4
1834 98.00 0.00 0.05 —
NH
3
597.1 155.38 0.00 0.07 —
Penicillin G 3420 0.00 334.00 — 0.16
CO
2
110.1 0.00 489.28 — 0.23
Water 1000 167.04 432.18 0.08 0.21
Biomass 1013 0.00 835.38 — 0.40
Total 2079.82 2090.84 1.00 1.00
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MATERIAL AND ELEMENTAL BALANCE 235
time dry air and 15 mol/min dry O
2
enters the chamber. The exiting gas stream contains 1%
(w/w) water. Draw and label the flow sheet and do the material balance for the humidifier.
Solution
The air humidification flow diagram for material balance is shown in Figure 9.2. The mass
flow rate for water into the humidification unit is:
The mass flow rate for dry oxygen is:
Basis of calculation: 1 min
Water Balance
Mass of water in is equal to mass of water out, since the enriched air with oxygen should
have only 1% humidity. Therefore the mass of humid air can be calculated as tie element.
25 0 01
250
4
4
.m
m 0 g Humid rich air with O
2
15
gmol
min
32 g
gmol
480 g/min O
2
1.5
litre
h
10 g
1 litre
1 h
60 min
25 g/min H O
3
2
Humidifier
O
2
480 g/min
Dry air
?
1% H
2
O
Liquid H
2
O
25 g.min
-1
1
4
32
FIG. 9.2. Flow diagram for oxygen enrichment and air humidification.
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236 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
Overall Material Balance
Total mass in is equal to total mass out.
The mass flow rate of dry air in is 1995 g/min.
Example 2
A fermentation broth containing Streptomyces kanamyceticus cells is filtered by a vacuum
rotary filter. The feed rate is 120 kgh
1
; each kilogram of broth contains 60 g of cells. To
improve filtration, filter aids are added at a rate of 10kgh
1
. The concentration of kanamycin
in the broth is 0.05%. The filtrate is collected at a rate of 112 kgh
1
. The concentration of
kanamycin in the filtrate is 0.045%. The filter cake contains cells, and filter aid is continuously
removed from the filter cloth.
(a) What is the moisture content in the filter cake?
(b) If the concentration of kanamycin in the filter cake is the same as in the filtrate, how
much kanamycin is absorbed per kilogram of filter aid?
Solution
Basis 1 hour operation.
The simplified flow diagram of continuous filtration with stream lines of material balance
is shown in Figure 9.3. Mass in is equal to mass out.
120 10 112 M filter cake
M filter cake 18 kgh
1
mmm m
m
m
123 4
3
3
2500 25 480
199
5 g
M
Filter cake
Filterate
112 kg/h,
0.045% kana
120 kg/h
6% cell
0.05% kanamycin
Filter aid
10 kg/h
Filter
FIG. 9.3. Flow sheet for continuous filtration.
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MATERIAL AND ELEMENTAL BALANCE 237
H
2
O Balance
Mass of water at inlet streams is equal to mass of water at the filterate and filter cake.
M (H
2
O in the filter cake) 0.79
The filter cake moisture content is calculated as:
The mass of water in the filtrate is 112 0.05 111.95 kg
Amount of kanamycin in the moisture 0.79 (0.00045)
3.6 10
4
kg
Kanamycin Balance
Mass of kanamycin in is equal to mass of kanamycin out.
0.0005 120 (112)(0.00045) (0.79)(0.00045) M
k.abs.
M
k.abs.
0.06 0.05075 9.245 10
3
kg
Kanamycin absorbed per kilogram filter aid
Various sources of carbohydrate are used in the fermentation processes. Molasses and corn
steep are the most common carbon sources used to generate energy for biosynthesis.
Having the correct composition and desired concentration is a necessary task in actual
experimental work.
Example 3
Corn steep liquor contains 2.5% invert sugars and 50% water. The rest of the feed is con-
sidered as residual solids. Beet molasses containing 50% sucrose, 1% invert sugar, 18%
water and remainder solids are mixed with corn steep liquor in a mixing tank. Water is
added to produce a diluted mixture with 2% invert sugar, 125 kg corn steep liquor and 45 kg
molasses, which is fed into an enzymatic hydrolysis tank.
(a) How much water is required?
(b) What is the concentration of sucrose in the final mixture?
9.245
10
9.245 10 kg/kg
4
0.79
18
100 4.4% moisture
120 1 0 06 0 0005[( . ) . ]
()
112.74 kg
112 1 0.00045 cake moisturee
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238 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
Solution
A flow diagram for diluting carbohydrates, corn steep liquor and molasses in the mixing
tank is shown in Figure 9.4.
Mass Balance
Let us assume mass in is equal to mass out.
125 45 W P
Now we can balance the invert sugar:
(0.025 125) (0.01 45) 0.02 P
P 178.75 kg
W 8.75 kg
The next step is to take sucrose into the balance:
(0.5)(45) (x)(P)
(0.5)(45) 178.75x
The concentration of sucrose in the product mixture is x 0.1259 or 12.59% of product.
The residual solids in corn steep, molasses and the diluted mixture are also balanced.
(0.31)(45) (0.475)(125) 178.75y
y 0.41
The amount of solid in the product mixture is 41%.
9.5.1 Acetic Acid Fermentation Process
In fermentation for the production of acetic acid, ethyl alcohol is used in an aerobic process.
In an ethanol oxidation process, the biocatalyst Acetobacter aceti was used to convert
ethanol to acetic acid under aerobic conditions. A continuous fermentation for vinegar pro-
duction was proposed for utilisation of non-viable A. aceti immobilised on the surface of
alginate beads.
P
2% Invert sugar
Sucrose x = 12.59%
Solid y = 41%
Mixing
tank
125 kg/h Corn steep liquor
2.5% Invert sugar
50% H
2
O
47.5% Solid
45 kg Molasses
1% Invert sugar
50% Sucrose
31% Solid
Water
w
FIG. 9.4. Flow sheet for diluting sugar and mixing tank.
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MATERIAL AND ELEMENTAL BALANCE 239
Example 1
The production rate of acetic acid was 2 kgh
1
, where the maximum acetic acid concen-
tration was 12%. Air was pumped into the fermenter with a molar flow rate of 200molh
1
.
The chemical reaction is presented in (E. 1.1) and flow diagram in Figure 9.5. Determine
the minimum amount of ethanol intake and identify the required mass balance for the given
flow sheet. The ethanol biochemical oxidation reaction using A. aceti is:
(E. 1.1)
The process flow diagram for fermentation of ethyl alcohol is shown in Figure 9.5. Assuming
steady-state condition and that the air intake has no moisture, volume and molar fractions
are about the same, ethanol is not going to vapourize while fermentation is taking place.
Assume the only biochemical reaction takes place is acetic acid production, which means
ethanol is not used for maintenance or ATP generation. No side reactions occur. Relative
molecular masses are: ethanol 46, acetic acid 60, oxygen 32, water 18 g/mol. Also, air consists
of 21% oxygen and 79% nitrogen.
Solution
The basis for calculation is 1 hour operation and 2 kg acetic acid produced.
Mass flow rate of air: (0.2kgmol/h)(28.84 kg/kgmol) 5.768 kgh
1
Mass flow rate of O
2
: 0.2 0.21 32 1.344 kgh
1
Mass flow rate of N
2
: 0.2 0.79 28 4.424 kgh
1
Sum of O
2
N
2
: 1.344 kgh
1
4.424 kgh
1
5.768 kgh
1
C H OH O CH COOH H O
Ethanol
25 2 3 2
A. aceti
æÆææ
Acetic acid
P, Product
2 kg/h, 12% acetic acid
16.67 − 2 = 14.67 H
2
O
Air
200 mol/h
E, ETOH, E kg/h
W, H
2
O, w kg/h
Fermenter
Off gas
G, kg/h
FIG. 9.5. Flow sheet for acetic acid production.
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240 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
The product stream with 12% acetic acid has a mass flow rate of:
Moles of acetic acid:
From the reaction stoichiometry, for each mole of acetic acid one mole of oxygen was used.
So the equal molar oxygen consumption is:
(3.333 10
2
)(32) 1.067 kg O
2
consumed
Excess amount of O
2
1.344 1.067 0.277 kgh
1
The off gas stream consists of all inlet nitrogen plus excess oxygen.
Off gas, G is the sum of excess oxygen plus inert nitrogen.
G N
2
excess O
2
4.424 0.277 4.701 kg
Off gas composition is 0.277/4.701 0.059, 5.9% O
2
and 94.1% N
2
Water Balance
The sum of water at the inlet stream and water generated by the chemical reaction is equal
to water at outlet stream.
H
2
O in H
2
O generated H
2
O out
Mass of H
2
O generated: (3.333 10
2
)(18 kg/kgmol) 0.6 kg
Water at inlet stream: 14.67 kg (0.6 kg, H
2
O generated) 14.07 kg
Ethanol Balance
Ethanol at concentration at the inlet plus ethanol generated is equal to ethanol in outlet
stream as unreacted reactant plus ethanol consumed. For each mole of acetic acid, one mole
of ethanol was utilised. The mass of ethanol used up is:
Mass of ETOH consumed (3.33310
2
)(46 kg/kgmol) 1.533 kg ETOH
Assume all the ethanol (100%) was taken by the bacteria and that there was no ethanol gen-
eration or evaporation. This means ethanol in is equal to ethanol consumed in oxidation
process.
Ethanol in ethanol consumed 1.533 kg ETOH
2 kg
60 kgmol/kg
3.333 10 kgmol
2
P
2
012
16 6
.
. 7 kg h
1
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MATERIAL AND ELEMENTAL BALANCE 241
9.5.2 Xanthan Gum Production
Example 1
Xanthan gum is produced using Xanthomonas campestris in a batch culture. Laboratory
experiments have shown that for each gram of glucose utilised by the bacteria, 0.23 g oxy-
gen and 0.01 g ammonia are consumed while 0.75 g gum, 0.09 g cells, 0.27 g gaseous CO
2
and 0.13 g H
2
O are formed. Other components such as phosphate can be neglected.
A medium containing glucose and ammonia dissolved in 20,000 litres of water is pumped
into the CSTR fermenter and inoculated with X. campestris. Air is pumped; the off gas
recovered during the entire batch is 1250 kg. The final gum concentration is 3.5wt%. What
is the percentage of excess air, and how much glucose and ammonia are required?
Reaction stoichiometry per unit mass of glucose:
Solution
Figure 9.6 shows the schematic diagram for production of xanthan gum with inlet and out-
let streams.
Overall Mass Balance
F A 1250 P
The amount of glucose required is:
1 kg glucose
0.035
0.75 g gum
0.0467 kg glucose
P
P
Ê
Ë
Á
ˆ
¯
˜
1 g glucose 0.23 g O 0.01 g NH 0.75 g gum 0.09 g cells
0
23
..27 g CO 0.13 g H O
22
P, Product
35% Xanthan gum
Air
A kg
Feed, F
20,000 kg H
2
O
Fermenter
Offgas
1250 kg/h
Mole fraction Mass fraction
0.21 0.233
0.79 0.767
F
IG. 9.6. Flow sheet for xanthan gum production.
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The mass oxygen based on reaction stoichiometry is:
The amount of ammonia needed is:
Products
Mass of cell with respect to mass of product stream is:
The mass of carbon dioxide generated is:
The amount of water according to reaction stoichiometry is:
O
2
Balance
The mass of oxygen in the inlet air stream is equal to oxygen consumed and the excess oxy-
gen leaving the outlet stream.
0.233A O
2
out 0.0107PO
2
used
The outlet stream
O
2
out 0.233A 0.0107P
N
2
Balance
Mass of nitrogen at the inlet is equal to mass of nitrogen in the outlet steam.
0.707A N
2
out
0.13 kg H O
0.035
0.75 g gum
0.00607 kg H O
22
P
P
Ê
Ë
Á
ˆ
¯
˜
0.27 kg CO
0.035
0.75 g gum
0.0126 kg CO
22
P
P
Ê
Ë
Á
ˆ
¯
˜
0.09 kg cells
0.035
0.75 g gum
0.0042 kg cells
P
P
Ê
Ë
Á
ˆ
¯
˜
0.01 kg NH
0.035
0.75 g gum
0.00047 kg NH
33
P
P
Ê
Ë
Á
ˆ
¯
˜
0.23 kg O
0.035
0.75 g gum
0.0107 kg O
22
P
P
Ê
Ë
Á
ˆ
¯
˜
242 BIOCHEMICAL ENGINEERING AND BIOTECHNOLOGY
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MATERIAL AND ELEMENTAL BALANCE 243
Mass of carbon dioxide generated is equal to mass of CO
2
at the outlet stream.
CO
2
out CO
2
generated 0.0126P
Off Gas Balance
1250 (0.233A 0.0107P) 0.767A 0.0126P
O
2
N
2
CO
2
1250 A 1.9 10
3
P
Mass of air is:
A 1250 0.0019P
20,000 0.04717P A 1250 P
Substitute A into the above equation, then solve for P:
20,000 (1250 0.0019P) 0.9528P 1250
P 20,949 kg
Now replace P in the above equation, then solve for A:
A 1250 0.0019(20,949) 1210.2 kg
Now solve for m
:
P 87.986 kg cells 263.96 kg CO
2
127.1 kg H
2
O
Off gas 57.8 kg O
2
928.2 kg N
2
263.96 kg CO
2
9.5.3 Stoichiometric Coefficient for Cell Growth
Stoichiometric coefficients for cell growth for the production of SCP from hexadecane is
given by the following reaction:
(9.9)
where CH
1.66
O
0.27
N
0.2
represents the chemical composition of cell biomass.
Example 1
Based on the stoichiometric relation of the above equation for SCP production, if RQ
0.43, find the stoichiometric coefficients a, b, c, d and e in (9.9).
CH O NH CH O N CO HO
Substrate Biomass
16 34 2 3 1.66 0.27 0.2 2 2
ab c d e
(SCP)
E
xcess air
57 8
224 15
100 25 8
57 8
0 0107 20949
100
.
.
.%
.
.( )
m
20,988.2 kg
mP (. . )/ ,0 0467 0 00047 20 000
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